Finding if exist the derivative at $x=0$ $f(x) =
\left\{
 \begin{array}{ll}
  \ln(\sin(x))-\ln(x) & \mbox{if } 0<x<\pi \\
        0 \\
  \ln(-\sin(x)) - \ln(-x) & \mbox{if } -\pi<x<0
 \end{array}
\right.$
The function is continuous, I've already checked that out. My problem is that I can not find whether the derivative exists at $x=0$.
Using the derivative by definition i get $$\lim_{h\to 0} \frac{\ln(\sin(h))-\ln(h)}{h}$$ and find that difficult to resolve. Thanks
 A: HINT
Note that
$$\frac{\ln(\sin h)-\ln h}{h}=\frac{\ln\left(\frac{\sin h}{h}\right)}{h}=\frac{\ln\left(1-\frac{h^2}6+o(h^2)\right)}{-\frac{h^2}6+o(h^2)}\frac{-\frac{h^2}6 +o(h^2)}{h}$$
and recall that as $x\to 0$ by standard limit $\frac{\log(1+x)}x \to 1$.
A: Justify the following steps:
$$\lim_{h\to0}\frac{\log\frac{\sin h}h}{h}\stackrel{\text{L'Hospital}}=\lim_{h\to0}\frac h{\sin h}\cdot\frac{h\cos h-\sin h}{h^2}=$$
$$=\lim_{h\to0}\frac{h\cos h-\sin h}{h\sin h}\stackrel{\text{L'Hospital}}=\lim_{h\to0}-\frac{h\sin h}{\sin h+h\cos h}=$$
$$\stackrel{\text{L'Hospital}}=\lim_{h\to0}\frac{\sin h+h\cos h}{2\cos h-h\sin h}=\frac0{2-0}=0$$
A: Since you have proved that $f$ is continuous at $0$, you can try and see whether
$$
\lim_{x\to0}f'(x)
$$
exists finite: in this case it will be the derivative at $0$ (possibly the most relevant theoretical application of l’Hôpital).
Note that $f$ is obviously differentiable for $x\ne0$, with
$$
f'(x)=\dfrac{\cos x}{\sin x}-\dfrac{1}{x}
$$
Now the limit is easier:
$$
\lim_{x\to0}\frac{x\cos x-\sin x}{x\sin x}=
\lim_{x\to0}\frac{x\cos x-\sin x}{x^2}=
\lim_{x\to0}\frac{-x\sin x}{2x}=0
$$
